Problem 87 - Chemistry: The Central Science

Question

Under appropriate conditions, molybdenum emits Xrays that have a characteristic wavelength of \(0.711 \AA\). These X-rays are used in diffraction experiments to determine the structures of molecules. (a) Why are X-rays, and not visible light, suitable for the determination of structure at the atomic level? (b) How fast would an electron have to be moving to have the same wavelength as these X-rays?

Step-by-Step Solution

Answer

X-rays are suitable for determining atomic structures due to their short wavelengths (0.711 Å) which are comparable to the size of atoms and the distances between them. This allows X-rays to resolve atom positions when diffracted by atomic planes in a crystal lattice, unlike visible light which has much longer wavelengths. Using de Broglie's equation, an electron would need to be moving at a speed of approximately \(1.020 \times 10^{6} m/s\) to have the same wavelength as these X-rays.

Step 1: Part (a): Suitability of X-rays for determining atomic structures

X-rays are suitable for determining atomic structures because of their short wavelengths. The characteristic wavelength of the X-rays in this exercise is \(0.711 \AA \) which is comparable to the size of the atoms and the distances between them. This allows X-rays to resolve atom positions when they are diffracted by the atomic planes in a crystal lattice. Visible light, on the other hand, has much longer wavelengths (typically between 400 to 700 nm) and thus is not suited for resolving atomic level structures.

Step 2: Part (b): Finding the electron velocity

To find the velocity of an electron with the same wavelength as the X-rays, we'll use the de Broglie's equation: \[ \lambda = \frac{h}{p} \] where \(\lambda\) is the wavelength, \(h\) is Planck's constant (\(6.626 \times 10^{-34} Js\)), and \(p\) is the momentum of the particle. For an electron, the momentum is given by the product of its mass (\(m_{e} = 9.109 \times 10^{-31} kg\)) and its velocity (\(v\)). First, we need to convert the given X-ray wavelength from Angstroms to meters: \[0.711 \AA = 0.711 \times 10^{-10} m\] Now we can solve for the electron velocity: \[v = \frac{h}{\lambda m_{e}}\] Using the given wavelength, Planck's constant, and the mass of an electron, we get: \[v = \frac{6.626 \times 10^{-34} Js}{(0.711 \times 10^{-10} m)(9.109 \times 10^{-31} kg)}\] Calculating the value, we obtain the electron velocity: \[v \approx 1.020 \times 10^{6} m/s\] Hence, an electron would need to be moving at a speed of approximately \(1.020 \times 10^{6} m/s\) to have the same wavelength as these X-rays.

Key Concepts

de Broglie's Equation

The concept of wave-particle duality is central to quantum mechanics and is ingeniously bridged through de Broglie's equation. This formula connects the wavelike and particle aspects of elementary entities like electrons. According to Louis de Broglie, every particle has an associated wavelength given by the equation: \[ \text{wavelength } (\lambda) = \frac{h}{p} \] where \(h\) represents Planck's constant and \(p\) symbolizes the momentum of the particle.

For example, when we talk about an electron behaving as a wave, it's the de Broglie wavelength that we refer to. In the context of our exercise, we use this equation to calculate the velocity of an electron that would result in it having a wavelength of \(0.711 \text{\AA}\), identical to that of the characteristic X-rays emitted by molybdenum. Thus, the equation allows us to draw parallels between the microscopic (electron's motion) and the macroscopic (X-ray wavelength) in a powerful yet simplistic fashion.

Determination of Atomic Structures

X-ray diffraction is a formidable tool in chemistry for determining the atomic and molecular structure of a crystal. When a beam of X-rays strikes a crystalline material, it is scattered in many specific directions. By measuring the angles and intensities of these diffracted beams, a crystallographer can produce a three-dimensional picture of the electron density within the crystal.

This image, when interpreted via mathematical models, unravels the positions of the atoms in the crystal, their chemical bonds, and other valuable structural information. The success of X-ray diffraction in resolving atomic structures relies on the fact that the X-ray wavelength (like the \(0.711 \text{\AA}\) wavelength from molybdenum used in our problem) is on a similar scale to the distances between atoms, allowing for the precise probing of the crystal lattice. This technique forms the backbone of modern structural chemistry and materials science.

Wavelength of X-rays

X-rays are a form of electromagnetic radiation with wavelengths typically ranging from 0.01 to 10 angstroms (\(\text{\AA}\)). They are shorter in wavelength than ultraviolet rays and longer than gamma rays. The relatively short wavelength of X-rays is crucial for their ability to discern structures at the atomic and molecular levels.

When we talk about the characteristic wavelength of an X-ray, as mentioned in the original exercise, we're referring to a specific wavelength that is emitted by an element (like molybdenum) when its electrons transition between energy levels. This distinct wavelength can serve as a 'signature' for the element and is leveraged for various applications, including identifying materials and probing their internal structure via X-ray diffraction techniques. Since the wavelengths of X-rays are comparable to interatomic distances, they can be used to construct highly detailed images of the atomic arrangement within crystals.